Free module implies trivial Tor Let $A$ be a commutative ring. If $M$ or $N$ aree free $A$-module then $Tor_{n}^{A}(M,N)=0$.
Since $Tor_{n}^{A}(M,N)=Tor_{n}^{A}(N,M)$ it suffices to deal with the case say when $N$ is flat right?
Take a projective resolution of $N$:
$... \rightarrow P_{n} \rightarrow P_{n-1} ... \rightarrow P_{0} \rightarrow N \rightarrow 0$
The next step is to tensor this sequence with $M$ yes?
$... \rightarrow P_{n} \otimes_{A} M \rightarrow P_{n-1} \otimes _{A} M \ .... $
But $M$ is free so by definition $M$ is isomorphic to a direct sum of copies of $A$. Hence:
$M \cong \oplus_{i \in I} A$ and since $P_{n} \otimes_{A} (\oplus_{i \in I} A) \cong \oplus_{i \in I} (P_{n} \otimes_{A} A) \cong \oplus_{i \in I} P_{n}$ we obtain:
the map $f_{n} \otimes 1:  \oplus_{i \in I} P_{n} \rightarrow \oplus_{i \in I} P_{n-1}$.
Now we need to consider:
$ker(f_{n} \otimes 1)/Im(f_{n+1} \otimes 1)$ yes? Why this quotient is trivial? Can you please explain?
 A: *

*If $M$ is a projective module, then there is a very convenient projective resolution for $M$, namely $$\cdots\to0\to0\to0\to M$$ If you tensor it by any module $N$ and compute homology, then obviously $\mathrm{Tor}_p(M,N)=0$ for all $p>0$.

*Alternatively, suppose that $N$ is any module and consider a projective resolution $$\cdots\to P_2\to P_1\to P_0$$ for $N$. Now, if $M$ is a projective module, then it is in particular flat, so if we tensor the projective resolution of $N$, which is an acyclic complex, with $M$, then the result will again be acyclic. It follows immediately that $\mathrm{Tor}_p(M,N)=0$ for all $p>0$.
A: The original projective resolution $\cdots\xrightarrow{f_{n+1}} P_{n} \xrightarrow{f_n} P_{n-1} \cdots \xrightarrow{f_1} P_{0} \xrightarrow{f_0} N \rightarrow 0$ was (by definition) an exact sequence, i.e. $\ker(f_n)/\text{im}(f_{n+1})=0$ for all $n$, i.e. $\ker(f_n)=\text{im}(f_{n+1})$ for all $n$. 
Now note that $\ker(f_n\otimes1)=\oplus_{i\in I}\ker(f_n)\subseteq \oplus_{i\in I}P_n$, and similarly for $\text{im}(f_{n+1}\otimes1)$.
